Question

The least value of $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$ for positive values of $x,y,z$ such that $x+y+z=9$ is
1). $0$
2). $1$
3). $2$
4). $4$

Answer 1

To get the answer of this question you must know the concept of Arithmetic mean and Geometric Mean and the relation between them. After that substitute the values as per the definition of AM and GM. Then simplify the inequality and you will get your answer. Now try the problem.

Complete step-by-step solution:
To get the solution of this question firstly you should understand the arithmetic mean and the geometric mean. And you should know the relation between both the means. After knowing these two concepts you can easily solve the given question. Let us try to understand these concepts.
We can calculate the Arithmetic Mean by taking the summation of all the terms present in the series divided by the total number of terms in the series.
Let us consider the given series is ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},.....,{{x}_{n}}$ then,
$AM=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+....+{{x}_{n}}}{n}$
We can calculate the Geometric Mean by taking the ${{n}^{th}}$root of the product of all the terms given in the series. Mathematically it is represented as
Let us consider the given series is ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},.....,{{x}_{n}}$ then,
$GM=\sqrt[n]{{{x}_{1}}.{{x}_{2}}.{{x}_{3}}.{{x}_{4}}......{{x}_{n}}}$
And we know that the relation between Arithmetic Mean and geometric mean is that AM is always greater than or equal to GM. So we can represent this statement as,
Now as per the definition of AM and GM, we get as
$AM\ge GM$
Now as per the definition of AM and GM, we get as
$\dfrac{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}{3}\ge \sqrt[3]{\dfrac{1}{x}\times \dfrac{1}{y}\times \dfrac{1}{z}}$
Multiply by $3$on both sides, we will get
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 3\sqrt[3]{\dfrac{1}{x}\times \dfrac{1}{y}\times \dfrac{1}{z}}$
There is the inverse relation so for minimum value of \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 1$$$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} the value of $x,\text{ }y\text{ }and\text{ }z$should be maximum. And it is possible if and only if $x=y=z$. And it is given that $x+y+z=9$.
So by substituting the value $x=y=z$, we will get the value of $x,\text{ }y\text{ }and\text{ }z$ as $3$.
Substitute the value of $x$, $y$ and $z$ in the right part of the inequality, we will get
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 3\sqrt[3]{\dfrac{1}{27}}$
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 3\times \dfrac{1}{3}$
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 1$
The value for $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 1$ is always greater than or equal to $1$. So the minimum value for the given equation is $1$.
Hence we can conclude that option $(2)$ is correct.

Note: Mean is basically used to find the central value of the given data. Median is also another way of finding the central value of the given data but its advantage is that it can also be used for the larger set of data whereas when mean is used to find the central value for the larger data then there might be a calculation error.